Distance covered in an infinitesimal time interval equals the magnitude of displacement during that interval :

$ds=|d\vec r|=\sqrt {64\cos^2 t+289\sin^2 t+225\cos^2 t} dt=17dt$

So the total distance covered during the given time interval is

$\triangle {s}=17(2020-2004)=\boxed {272}$ m.

Displacement of the particle during this interval is $33.638$ m., which is approximately $8.086$ times smaller than the distance covered.

We have

$\vec{r} = (8\sin t + 55)\hat{i} - (17\cos t + 44)\hat{j} + (15\sin t - 54)\hat{k}$ $\frac{d\vec{r}}{dt} = \frac{d}{dt}(8\sin t + 55)\hat{i} - \frac{d}{dt}(17\cos t + 44)\hat{j} + \frac{d}{dt}(15\sin t - 54)\hat{k}$ $\vec{v} = (8\cos t)\hat{i} + (17\sin t)\hat{j} + (15\cos t)\hat{k}$ $|\vec{v}| = \sqrt{(8\cos t)^2 + (17\sin t)^2 + (15\cos t)^2}$

$\boxed{v = 17 m/s}$

So, distance $d$ travelled by particle is

$d = v\Delta t$ $d = (17)(2020 - 2004)$ $\color{#3D99F6}{\boxed{d = 272m}}$